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Differential equation question (HELP)

  1. Sep 17, 2006 #1
    Hi

    Given the non-homogenous higher order differential equation

    [tex]u'' - 2 u' - 8 u = 6 - 8t[/tex]

    with the intial conditions u(1) = 0 and u'(1) = 1.

    Find the solution.

    I'm told by my professor that I'm supposed to use the fact

    [tex]{u''} + {a_1} u' + {a_0} u = f [/tex]
    The characteristic polynomial for this differential equation is supposedly

    [tex]z^2 + {a_1} z + {a_0} z = 0[/tex]

    Using this fact I find root of the polynomial of the differential equation above to be [tex]\lambda _{1,2} = 4,-2[/tex]

    The I'm suppose to use the fact that (Existence and Uniqueness theorem of Non-liniear differential equations) if [tex]\lambda {_1} \neq \lambda {_2}[/tex]

    [tex]u_0 (t) : = \int \limit_{t_0} ^t \frac{e^{\lambda_1 (t-s)} - e^{\lambda_2 (t-s)}}{\lambda_1 - \lambda_2} f(s) ds[/tex]

    [tex]t \in I[/tex]

    My solution.

    Since lambda 1 and lambda 2 er different then there exist a unique partiqular solution in the form:

    [tex]u(t) = C_1 e^{6-8t} + C_2 e^{-8} = 0[/tex]

    By inserting part 1 of the condition into the above, I get that C_1 = C_2 = 0.

    initial condition 1 = true.

    I differentiate u(t) and get [tex]u'(t) = -8 e^{-2} C_1 = 1[/tex]

    here I get that [tex]C_1 = \frac{1}{-8{e^{-2}}} = \frac{-e^2}{8}[/tex] and this proves that u'(1) = 1

    Am I on the right track here now?

    Sincerley Yours
    Fred
     
    Last edited: Sep 17, 2006
  2. jcsd
  3. Sep 17, 2006 #2

    arildno

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    That would be the solution of the associated HOMOGENOUS problem (i.e, when f=0).
    In addition to that, you need a particular solution of your diff. eq, try with the trial function at+b, where a, b are constants to be determined
     
  4. Sep 17, 2006 #3
    Please look at my corrected solution above. Am I on the right track?

    Sincerely
    Fred
     
  5. Sep 17, 2006 #4

    arildno

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    Your u(t) is NOT a solution of your original diff eq; it is part of it!
     
  6. Sep 17, 2006 #5
    Okay the I not sure but then full solution is

    u(t) = C_1 e^{6-8t} + C_2 e^{-8} = 6 - 8t ??

    Sincerley

    Fred
     
  7. Sep 17, 2006 #6

    arildno

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    No, insert in your diff. eq. the trial function:
    [tex]u(t)=C_{1}e^{\lambda_{1}t}+C_{2}e^{\lambda_{2}t}+at+b[/tex]
    What equations do you see that "a" and "b" must fulfill in order for u(t) to be a solution?
     
  8. Sep 17, 2006 #7
    a and b must have such values that the solution is equal to the original differential equation ?

    /Fred
     
  9. Sep 17, 2006 #8

    arildno

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    Almost:
    a and b must have such values that the trial function is a SOLUTION of the original differential equation.
     
  10. Sep 17, 2006 #9
    by inserting a trail equation u(t) = at^2 + b

    I get a = 0 and b = 2.

    Does that sound right?

    Sincerely Fred
     
  11. Sep 17, 2006 #10

    arildno

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    No, you should insert at+b:
    the second derivative of this is zero, so you end up with the equation for a, b:
    [tex]-2a-8(at+b)=6-8t[/tex]
    Thus, we get a=1 to match the first power in t, and to zeroth order in t, we're left with:
    -2*1-8b=6, that is b=-1
     
  12. Sep 17, 2006 #11
    Okay then the solution is

    [tex]u(t)=C_{1}e^{\lambda_{1}t}+C_{2}e^{\lambda_{2}t}+ t - 1[/tex]

    But what about the lambda values could you give me a hint on how til determain them?

    Sincerely

    Fred
     
  13. Sep 17, 2006 #12

    arildno

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    Well, you have already found them as the solutions of the characteristic equation!
    Surely you have seen this type of exercises before?
     
  14. Sep 17, 2006 #13
    Yes of course, I have had the flue this last week, so I have been under the weather ;)

    Just to recap lambda values are found as the solution of

    u(t) = z(t)^2 - 2 z(t) - 8 = 0 ?

    Sincerely Fred.

    p.s. Talking about existence and uniquness. The exist only one solution for the differential equation. Since the solution polynomial for diff. equation is unique ?
     
  15. Sep 17, 2006 #14

    arildno

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    Yes.
    You now find the the UNIQUE solution of the whole initial value problem by fitting the C's so that the initial conditions are satisfied.
     
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